Print Cube Root of A Number in C++

Confronting with numbers is the first thing each one of us has learned in school. However, as we step up high more and more difficult questions based on the same fundamentals seem to appear in our way. It’s the same with programming, using the same fundamentals to find solutions for multiple complex problems. Now the cool thing is, you can be creative in your way eventually coming up with a unique solution. And if you’re able to do this, then you’ve clearly understood how to use programming for others and your good. But keeping that aside, we’ve talked about numbers before. So what if we find the cube root of numbers, seems too easy right!. Well, So let’s quickly find out how to Print the Cube Root of A Number in C++.

With the help of this program, you will be able to find the root square of integers. Practising these types of questions also helps to get an upper edge in Competitive Programming.

What’s The Approach?

• We’ll use Binary Search Algorithm to find the cube root of the given input number n.

• Therefore we’ll initialize start=0 & end=n, and we’ll find the middle value in this list by executing (start + end)/2

• We’ll set an error point e, below which if the calculation goes will be set as the cube root of n. So the condition is (n – mid*mid*mid) < e

• However, if (mid*mid*mid)>n we’ll set end=mid. And if (mid*mid*mid)<n we’ll set start=mid

Also Read: Implement Linear Search using C++

C++ Program To Print Cube Root of A Number

Input:

n = 8

Output:

The cubic root of 8.000000 is 2.000000

// C++ program to find cubic root of a number

#include <bits/stdc++.h>
using namespace std;

// Returns the absolute value of n-mid*mid*mid
double diff(double n,double mid)
{
    if (n > (mid*mid*mid))
        return (n-(mid*mid*mid));
    else
        return ((mid*mid*mid) - n);
}

// Returns cube root of a no n
double cubicRoot(double n)
{
    // Set start and end for binary search
    double start = 0, end = n;

    // Set precision
    double e = 0.0000001;

    while (true)
    {
        double mid = (start + end)/2;
        double error = diff(n, mid);

        // If error is less than e then mid is
        // our answer so return mid
        if (error <= e)
            return mid;

        // If mid*mid*mid is greater than n set
        // end = mid
        if ((mid*mid*mid) > n)
            end = mid;

        // If mid*mid*mid is less than n set
        // start = mid
        else
            start = mid;
    }
}

// Driver code
int main()
{
    double n = 8;
    printf("Cubic root of %lf is %lf\n",
        n, cubicRoot(n));
    return 0;
}